Ask your own question, for FREE!
Mathematics 14 Online
OpenStudy (anonymous):

Determine whether the integral converges or diverges. Find the value of the integral if it converges.

OpenStudy (anonymous):

OpenStudy (anonymous):

@Reaper534 can u help?

terenzreignz (terenzreignz):

Let's just focus on the integral first, keeping in mind that... \[\Large \int\limits_1^\infty x^{-\frac43}= \lim_{b\rightarrow\infty}\int\limits_1^bx^{-\frac43}\]

OpenStudy (anonymous):

okay

terenzreignz (terenzreignz):

Well, can you integrate \[\Large \int x ^{-\frac43}=\color{red}?\]

OpenStudy (anonymous):

yes i get -3/3sqrt(x)

terenzreignz (terenzreignz):

This \[\Large \frac{-3}{3\sqrt{x}}\]?

OpenStudy (anonymous):

yes!

terenzreignz (terenzreignz):

Or this \[\Large \frac{-3}{\sqrt[3]x}\]

OpenStudy (anonymous):

the second one

terenzreignz (terenzreignz):

lol... it's cube root, not 3sqrt because \(\Large3\sqrt x\) means something entirely different, ok?

OpenStudy (anonymous):

Okay

terenzreignz (terenzreignz):

Okay, so, we have... \[\Large \int\limits_1^b x^{-\frac43}=\left.\frac{-3}{\sqrt[3]x}\right]_1^b\]

terenzreignz (terenzreignz):

Can you evaluate this bit?

OpenStudy (anonymous):

i think

terenzreignz (terenzreignz):

Well then, what do you get?

OpenStudy (anonymous):

hmm i'm working on it

OpenStudy (anonymous):

eh...i'm getting a wrong answer :/

terenzreignz (terenzreignz):

Fundamental theorem of Calculus? \[\Large \int\limits_a^b f'(x)dx = \left.f(x)\right]_a^b = f(b)-f(a)\]

OpenStudy (anonymous):

ohhhh okay

terenzreignz (terenzreignz):

So... \[\Large \int\limits_1^b x^{-\frac43} \ dx=\left.\frac{-3}{\sqrt[3]x}\right]_1^b=\color{red}?\]

OpenStudy (anonymous):

so i plug in b and 1 into that then subtract?

terenzreignz (terenzreignz):

Yes... but that's not the end yet, just plug in for now, and tell me what you get :)

OpenStudy (anonymous):

okay so -3/3sqrt(b) - -3/3sqrt(1)

terenzreignz (terenzreignz):

I'm assuming by 3sqrt you mean cube root :D Okay, that being the case, you're right :) \[\Large \int\limits_1^b x^{-\frac43}=\frac{-3}{\sqrt[3]b}+\frac3{\sqrt[3]1}= -\frac3{\sqrt[3]b}+3\] Catch me so far?

OpenStudy (anonymous):

yes

terenzreignz (terenzreignz):

Now, we're supposed to take the improper integral to infinity, right? Remember this... \[\Large \int\limits_1^\infty x^{-\frac43} \ dx = \color{red}{\lim_{b\rightarrow\infty}}\int\limits_1^bx^{-\frac43} \ dx\] Now is the time to apply that limit (which we haven't done yet) \[\Large \color{red}{\lim_{b\rightarrow\infty}}\left(-\frac3{\sqrt[3]b}+3\right) \]

OpenStudy (anonymous):

okay

terenzreignz (terenzreignz):

So... evaluating the limit...?

OpenStudy (anonymous):

umm i get 3? :/

terenzreignz (terenzreignz):

That's right!

OpenStudy (anonymous):

ohh okay lol

OpenStudy (anonymous):

so thats the final answer?

terenzreignz (terenzreignz):

Well, you technically have two questions, but since there was an answer, then the integral converges, and it converges to 3 ^.^

OpenStudy (anonymous):

okay thanks

terenzreignz (terenzreignz):

No problem :)

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!